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NATIONALADT@OKi-‘LYMvWIXEE. .

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FORAERONAU.HCS-.

TECHNICALNOTE

Ivo.

... ..—_

1014

—... .-....—._

AND.GENERAL IIIS!FABILII!YOIPMONOCOQUliCYLINDERS

WITH CUTOUTS

CALCWIIATION03’THE SZ!RESSXSIliA CYLINDER -

wITH A SYMM13TRICCUTOUT

J. Hoff, Bruno A. Bol’ey,and BertramPolytechnicInstituteof Brooklyn

NOTTOBETAKENl?TiOMTHISROOM

WashingtonJune 1946

...— —_,_ . . .. ... ___

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LANm,JIyMEXWJAL..+tzgtw&LAEORATC)RY‘- ““““::LarderFieicLVW ,..

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https://ntrs.nasa.gov/search.jsp?R=19930085240 2020-04-04T17:31:35+00:00Z

NATIONALADE3.$(2RYCOMMITTEEFOR AERo~AuTIcs-.—

TECHNICALNOTE NO. 1014

STRESSES IN AND GENERAL INSTABILITYOF MONOCOQUECYLINDERS

WITH CUTOUTS

XI - CALCULATIONOY THE STRESSES IN A CYLINDER

WITH A SYMMETRICCUTOUT

By N. J. Hoff, Brund A. Boley,and Bertram Klein

.

SUNMARY

A numericalprocedure is yresented for the calculationof the stresses in a monocoque cylinderwith a outout. In

~ theprocedure the structureis broken up into a great many..., units; the forces in these uhits correepondlngto specifiedw d.ietorti.onsof the unite are calculated;a set of linear

equationsis establishedexpressingthe equilibriumcon-ditions of the units in the distortedstate; and thesimultaneouslinear equationsare solved. A fully workedout numericalexample,correspondingto the applicatio-nofa pure bendingmoment, gave results in good agreementwithexperimentscarriedout earlierat the PolytechnicInstitute ------—of Brooklyn.

6INTRODUCTION

Actual airplanesdfffer greatly from the idealizedstructuresthat underlie most theoreticalanalyses.. Thereason for these deviationscan be found in the greatdifficultiesinvalvedin applyingthe theory of elasticityto the irregularand complex structuralparts of airplanes.It is believed that the most promisingapproachto these

a complexprobleme is the one in which the structureisimaginedto be %roken Up into a great uumher of ‘Units,”the forces in theseunits correspondingtc specifieddis-Atortionsof the units are calculated,a set ot”Siaearequationsis establishedexpressingthe equilibriumcon-ditions of the units in the distortedstate, and thesimultaneouslinear equationsare solved.

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NACA TN NO. 1014 2

The set of linear equations,excludingthe load terms,forms what SouthwelL(reference1) called the “operationstable.” In Southwell?srelaxationprocedurethe equationsare solved by a method of step-by-stepapproximations. Inthe past 2 years a considerableamount of work has been doneat the PolytechnicInstituteof Brooklyn in applyingSouthwelllsmethod to the stress analysis of reinforcedthin–walled structures. It was easilypossible to establisharapidly converging

7rocedure in the case of stiffenedpanels

(references2 and 3 . In the case of ring problems(refer-ences 4 and 5) the convergencewas found to be poor as arule, and suggestionswere made for solvingthe equationseitherdirectly by matrix methods, or by a proceduredenotedas the “growinguniti~’method,

The problem of the calculationof the stresses in areinforcedmonocoque cylindercombinesthe two elementsdiscussedin the earlierreports,namely, reinforcedpanelsand rings. The authorswere unable to devise a pure step-by-stepprocedurethat would lead to a solution of theequationsrepresentedby the operationstable with a reason-able expenditureof work and time. On the other hand, thesolutioncan be found comparativelyeas’ilyif the operationstable is set up with the aid of the expressionsdeveloped

kin this report and the equationswhich the operationstable

i representsare solvedby matrix methods. A fully worked‘w’ out numerical examplegave results in good agreementwith

the tests described in reference6.

This investigation,conductedat the PolytechnicInstituteof Brooklyn,was sponsoredby and conductedwiththe financialassistanceof the National Advisory Committeefor Aeronautics.

SYMBOLS

8 a distancebetweenrings measured in the x-direction

a,b,c,d,e,f,g,h,j,k,l,m,n,p,q,r,s,t,u,r’,s’, tf portions of operationstable

ansbntcn Fourier coefficients

A cross-sectionalarea of a stringerplus itseffectivewidth of sheet

-—

RA,CATN 2?0,1014 3

A,B,C,i3,E,F,G,H,J,K,O,P points UE Intersectionef rings and stringers

A,B,C,D,E,I?,G,H,J,K,L,L,N,P3Q,R,S,T,U,Y,VI,

A–A, B–B, C-C,D-D, At–A:,

. Bf_Bl, ~l_~f,Dt_Dt

AB, BC, CD,DDI

L

a

porttons of operat$on$ta%le

ehear stress

shear modti~us

moment of inertiaof ring cross sectionplusits effecttvewidth for bending in itsplane .-

distancebetweenstringersmeasured alongthe circumference,that is, developedlength of ring segment between adjacentstringers

bendingmoment acting in a transverseseotionof the cylinder .

index

bendingmoment acting in the plane of a ring

unknown coefficients.-.

resultantforce acting in a tr=ns?ersesectionof the cylinder

shear flow

radius of cylinder

NACA TN NO. 1014 4

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R

Ss

ii

T

u

v

w

x

an~ar9at

P

‘Y

r = Gt/2

n

radial force

line of symmetry

sheet thickness

tangentialforce

tangentialdisplacementof an intersectionpointof Q ring and etringer

radial displacementof an Intersectionpoint ofa ring and stringer

rotation of a sectionof”a rtng in its own ylane

axial force

coefficientsused In the calculationof forces--—

and moments causedby the shear flow existingin a panel

angle subtendedby ring segment

section—lengthparameter

vertical downwardtranslationof a ring as a rigid.-.—

body— —

--—

A = GtL/a

f axial displacement;also the ratio of effectiveshear area of a section to the actual areaof the section

Q angular coordinate

u) rigid body rotation of a ring—-— -—

c1= Gta/4L

The symbolsused tO d.e~oteinfluencecoefficientsaredefined in the followingmanner:

(~b) stands for the force or moment a caused by a unitmovement in the directionof b (which direction3.sthat of

..

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NACATNNo.1014 5-——

theforceR or T, orofthemomentN). Thus ~) Isthemomentduetoa unitrotation;while (%?) isthetangentialforcearisingfroma unitradialdisplacement.Further,tadistinguishthereactionsat thefixedendfrom%hoeeat themovableeti~thesubscriptsF and M areemployed.Ccmse-quently,(nt)F isthemoment~isingatthertiedendofthecurvedbarasa resultofa tit tangentialdisplace-mentofthemovableend;while (~)~~st~dSforthetan.‘gentialforceat themovableendduetoa un&ttangentialdisplacementofthatend.

Itshouldbenotedthatallthereactionsconsideredhereareactingfromthesupportuponthecurvedbar.

CAHXILATIONOF.TEEDEFORMATIONSANDSTRESSES

SIMPL13KIEDCYLINDER

AsswptlonsRegardingtheStructure

INTHE

Theactua~monocoquecy~er (fig.~) contains16stringersand8 rings,includingthe2 endrings.Thecut-outinthecylinderextendsoverthreeringfieldsandtwostringerfields.Thereare,therefore,106panelsinthestructwenotcountingthe6 panelscutout. Sincethe%nit” ofthestructureisthepanelanditsborderingstringerandringsegaents,itappearsthattherearetoo_ wits tOp-t a calculationofthedisplacementandstresseswitha reasonableamountofwork. Henceitwasdecidedthattheactualstructureshouldbe replacedby thesimplifiedoneshowninfigurelb.

Itwasanticipatedthattheeffectofthecutoutuponthestressdistributionwouldbe importantonlyintheneighborhoo~ofthecutout.Thisisthejustif~a.tionforchoosingsmallsizetits close*Oandlargesizeunitsfaxtherawayfromthecutout.- ordertoavoida changeinthetotalamountof stringercrosssectioninthecylinder,thesizesofthestrtngersborderinglargepanelsweretncreased,asmayte seenfrmnthedatacontainedinfigure1%. Thetotalwidthof sheetletweenadjacentstringerswasconsideredeffectiveincarryingnormalstressintheaxialdirectionandwasdistributedevenly%etweentheadjacentstringers.Becauseof thtsasstnnptionthe

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lT4CAThiNo, 1014

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results of the calculationscan be anticipatedto agreewith experimentonly when the loads appliedare so smallthat the panels of sheet are in an unbuckled state. Whencalculationsare to be carried out for higher loads, asmallervalue should be ~ssumedfor the effectivewidth.The effectivewidth of sheet actingwith the ring wastalcenequal to the wiLth of the ring.

Because of the symmetryof both structureand loadingonly one-quarterof the monocoque cylinderwas consideredin the calculations.

—

In regard to the mechanicalpro~ertiesof the elementsof the structurethe followingassumptionswere made:

1. The stringershave only extensionalrigidity;they—--.—

are very weak in bending and torsion.

2. The rings are resj.stantto shearingand bending intheir plane of curvatureand to extension,but are veryweak in bending out of their plane and in torsion.

3. The sheet is resistant only to shearingdeformations.The extensionalrigidity of the sheet is taken into accountthroughthe assumptionof an effectivewidth the area ofwhich is added to the cross-sectionalarea of stringersand rings. A oonseauenceof this assumptionis that theshear stress must be constantover any single panel.

These assumptionsare believedto representthe essentialfeatures of the elementsof the structure. The agreement -betweenthe results of the calculationsand the experimentssubstantiatesthis belief. -—

-~

The Unit Problem.—

1. Axial tiisplacement& at point A.— When point A

in ftgure 2 is d~splaceda distance ~ axially(in the x- “direction),the force ‘D str exerted by the stringersegmentAD upon the constraintat D is

‘D str = (EAstr/a)L (la)—-——

EACA 7?NNcJ.1014 7

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where ‘str is the cross-sectionalarea of the stringerplus its effeattvewidth of sheet. At the same time aconstantshear flow q~~ will be oaused in the panel:

.

‘Atj= (Gt/2L)~ (lb)

In figure 2 the shear flow 4s tndicatedas acting from thesheetupon the edge reinforcements. As in reference2 one-half the shear force transmittedfrom the sheet to the edgestringeris consideredto be actingupon each of the con-straintsat the end points of the stringer. Consequently,the axial force ‘D sh transmittedby shear to the con-straintat D is

x~ s~= -(Gta/4L)g (lC)

!l?hetotal axial force x~ actingupon the aonstratntatD is

‘D=[(EA6tr/a)- (Gta/4L)~~ (1)

!lheshear flow qA actingupon the ring segment ABcausesforces and moments to act upon the constraintsatA and B which can be computedfrom the data give~ inreference5. It should be remembered,however,that inthis referencethe quantities listed are forces and momentsactingfrom the constraintsupon the ring segment at theend point toward which the shear flow is directed,and thesighs correspondto the beam convention(fig. S), At theother end the same numericalvalues apply, %ut the signs areinverted. Since in figure 2 the shear flow is directedto-ward point A, and since the forces and moments acting uponthe constraintare sought, a multiplicationby –1 must becarriedout. The forces and moments actingupon the con---—straintat A are, therefore,accordingto the beam con-ventIon

~A=- at .~qk = - at(Gt/2)t

RA = - ar LqA = - mr(Gt/2)~)

(2a)

--—

NA=C-CLn L2qA =- c+( Gt /2)E J

NACA TN NO. 1014 8

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Similarlyat point B

TB = at LqA = at(Gt/2)~

RB = a= ‘qA = ~r(Gt/2)~

1

(2b)

‘B = an IJaqA= anL(Gt/2)f

If these equationsare rewrittento agree with theframe convention(fig. 3), the following iS obtained:

H =cct(Gt/2)E

‘A = -ar(Gt/2)~

%= a#!Gt/2)t

‘B = ctt(G@)k

N3 = anL(Gt/2)g J

(2)

(3)

Since the shear flow transmittedto ring segment CDis equal and opposedto that transmittedto ring segmentAB,

!?=-92c B

TD = -TA

‘C = ‘RB

RD = –RA

‘C = ‘NB

}(4)

ND = -NAJIt is easilyYerified’onthe basis of the principles

developedin reference2 that the axial forces actinguponthe constraintsat points A, B, and C are:

XA = -[@A,Ja) +(~ta/4L)I~ (5)—-.—

XB = (Gta/4L)~ (6)

xc = (Gta/4L)E (7)

.

w

b

2. TangentialDisplacementat point A.- When po~nt Ais displacedtangentiallythrougha distance u (in the pos-itive directionof the beam convention)the forces and momentsactingupon the ring segmentat A and B can be computedfrom the data presentedin reference5.

n

‘A = tt~u

n

‘A = rtMu

n

‘A = ntMu

(aa)

TB = ttru

%3= rtru

:]

(ah)

‘B = ntFu

At point A a positive u displacementaccordingtothe frame conventionis equal to a negative u displacementaccordingto the beam convention. If a positive u dis-placement is now assumed accordingto the frame convention,the forces and moments are exyresseclas they are actingupon the constraint,and the signs are written in agree-ment with the frame convention,equations(~a) and (8b)become —. .

-

‘B = ttTu

n

‘B = rtru

(8)

.-.

(9)n

‘B = ntru J

NILCATN HO. 1014 10

At the same time the displacernentiu gives rise to auniform shear flow in the panel. The directionof the shearflow aotimg upon the edge reinforcementsof the panel isindicatedin figure 4. Its magnitude can be calculatedwiththe aid of Maxwell~sreciprocaltheorem. When A was dis-plaoed in the x-direction,the inducedshear flow caused atangentialforce at A of s magnitude

‘A = mt(Gt/2)~

accordinqto the first of equations(2). Hence when A Isdisplacedtangentiallya distance u, accordingto thereciprocaltheoremthe axial force causedat A must heequal to

XA = at(Gt/2)u (10)

This axial force, however, can occur only If themagnitudeof the shear flow q&u is given by .

qA~ = Iat(Gt/a)uI (ha)

The directionof the shear flow must be determinedfromequation(10). Since at is negativewithin the range ofangles representedin the graphs of reference5, the forceexertedupon the constraintat A is directed toward the leftin figure 4. Consequently,theehear flow acting upon theedge reinforcementsof the panel must be as shown in figure4.

It may be seen that the directionof the shear flow isthe same in figures 2 and 4. Equations(2) and (3) can beeasily transformedto correspondto the values causedbythe tangentialdisplacement.

-u~(GtL/a)u

arat(GtL/a)u

–anat(GtLz/a)u1

(11)

NACA TN NO. 1014.

d

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‘B = -@GtL/a)u

RB = -a,rmt(GtL/a)u

‘B = -~.+ GtL2/a)u

Equations(4) are valid here also

1 (12)

The total forces and moments at A are obtainedby addi-ng up correspondingvalues in equations(S) and (11):

‘A = <~tk + a~(GtL/a)]u1

RA . [Tt~+ CCrmt(ML/a)}u

‘A = -[=11 + V,nat(GtL=/a)]u

S5niLarlythe total forces and moments atcalculatedfrom equations(9) and (12):

‘B= [+_ m~(GtL/a)]u

RB = [zF- fi.rcct(GtL/a)lu

NB= [GT - tinctt(GtLa/a)~

At C the forces and moments are:

‘c = a,~(GtL/a)u

‘c = arat(GtL/a)u

‘c = an@tLa/a)u

At D the forces and moments are:

‘D = a~(GtL/a)u

RD.- arut(GtL/a)u

‘D = anc@GtLa/a.)u

.-._ _

(13)

be

(14)

(15)

(16)

NACA TN NO. 1014 12.

The axial forces can he calculatedfrom the followingequa-~ tions:

‘B = xc = -xD = -Xk = -c@t/2)u (17)

~ Radial Displacement v and Rotation w at Point A.

The calculationsneeded for determiningthe forces and mo-ments correspondingto these distortionsare quite similarto those given under (2). The results of the calculationsare presented in the diagramsto be discussedunder (4).

.. The four-panelA problem.- In the general case any

point belongs simultaneouslyto four differentpanels. Adisplacementof the point, therefore,will cause forces andmoments to appear in four panels. These forces and momentscan be calculatedwithout difficultyfrom the results of thesingle-panelunit problem. For the convenienceof the stressanalyet the four-panelproblemhas been worked out and theresults of the calculationsare presented in figures ?’to 10.In figure 10a the sign conventionis shown.

lYhenthe sheet is in a buckled state in any particular-—— _

panel, a reducedvalue ehouldbe used for G. If one or twoof the four panele adjacentto a point are cut out, G shouldbe put equal to zero for those panels.

The OperationsTable

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The operationstable containsthe forces and momentsactingupon the constraintscaused by the individualunitdisp3,acements. The individualitems in the table are calcu-lated accordingto the principlespresente~ in the precedingsection. Each number in the operationstable representsthevalue of the force quantity indicatedat the left end of therow in which the number is located, causedby the unit dis-placement indicatedat the top of the column in which thenumber is located. This arrangementdiffers from the oneused in references2 to 4 insofaras the headings of thecolumns in this table are those of the rows in the referencesmentioned,and vice versa. The sets of the headings of thecolumnsand rows, however, are interchangeableeince theoperationstable is eymmetricwith respect to its principaldiagonal. The number of individualoperations- that is,the number of degrees of freedom of the structure- can beileterminedwith the aid of the followingconsiderations.

NACA TN No. 1014” 13.

As was stated ea~lter,it sufficesto consideronlyd one-quarterof the entire structurein the calculationsbe-

cause of the symmetry of both structureand loading.

Points K, L, M, H, and O of the end ring are consideredrigidlyfixed. In the tests described in reference6 theywere attachedto a rigid frame. Points B, C, D, Y, G, andH are free to move axially,tangentially,and radially,and

. the sectionsof the rings at these points are free to rotatein the plane of the ring. Points A, E, and J are free tomove axially and radially,their other two types of motion

. being excludedbecause they are antisymmetricwith respectto the plane of symmetry of the cylinderpassing throughpoints K, A, O, E, and J.

Altogether,the systemhas 30 degrees of freedom. Thedisplacementscorrespondingto them are arrangedacoordingto the followingscheme in the operationstable (table 1).Yirst, all the axial displacementsare listed, nine innumbdr. They are followed by aZl the other displacementsof each of the points of rfng ABCDE, arrangea in the order

. of tangentialdisplacement,radial displacement,and rotation.Al%~gether,there are 11 such operationsif the antisymmetricdistortionsat points A and 3 are excluded. Finally,the 10

% individualoperationsin the plane of ring 3’GHJare listed.

In t!?.eappendix it iS shown by means of typical exampleshow the entries in the operationstable are calculated.

Calculationof Displacements

As far as the loading is concerned,the forces actingin the end sectionsupon the individualstringersare notstipulated,but it iS reauiredthat they add up to a purebendingmoment acting in the vertical axial plane of symmetryof the cylinder. On the other hand,it is known that ringFGILImust r’emainplane during the distortions. The calcu–latiioaof the distortionsof the structureis, therefore,carriedout in the followingmanner.

Plane FGHJ iS assumed to be rotatedabout the horizontal. axis in its plane passing throughpoint H. The angle of

rotation is defined by the assumptionthat the axial dis-. placementof point J is 0.001 inch.

It was shown in reference4 that the operationstable# representsa set of simultaneouslinear equations. Eor

instance,the first row in the operationstable (table 1)may be written in the form

-113.9~A + 38.9~B+ 0.934vA+ 11.9uB- 0.934vB + 0.305wB= O

● As a second example,row 5 reads:

17tD - 71953 + ?.3g= + 394.7tJ

- 11.7uH– 3.78vH- 5.05wH -I-3.78vJ = O

In this equationthe values of ~H and ~J are O and 1,respectively,accordingto the assumptionsmade regardingtherotation of ring YGHJ. (In the operationstable the unitaxial displacementis 0.001 In.) Consequently,the entryin column 9 and row 5 is a known quantityhaving the value‘- 394.? x 1 = 394.7 pounds. It can be taken over to the right- -hand eide of the equation. Since the entry In column8 androw 5 is zero because t~ = 0, the equationcorresponding%to row 5 may be written as

17~D - 719~E- 11.7uH-.3.78vH- 5.05wH+ 3,78vJ =-394.7

>

.

It is easy to see that the completeset of equationscorrespondingto the assumedrotation of ring FGHJ can beobtainedby multiplyingthe figures listed in each columncorresponding=to an assumed axial displacementby the assumedvalue of the displacement,transferringthe numbers obtainedto the right-handside of the equations,and addingupalgebraicallythe numbers on the right-handside of eachequation, .

The terms containedin rows 6 to 9 of the operationstable add up to the axial forces actingupon points F, G, H,and J, respectively. They need not be equatedto zero since,because of the symmetry,equal and oppositeforces originatingfrom the omitted other half of the cylinderautomaticallybalance them. For this reason, rows 6 to 9 must be omittedfrom the set of equationsto be solved..They will be usedlater for establishingthe nature of the externalloadingof the cylinder.

NACA TN HO. 1014 15.

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%

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The set of the remaining26 equationswas solvedbyDoolittleismethod (references7 and 8). In other words the26 displacementsand rotationswere calculatedthat correspondto zero resultantforce and moment at eaoh point of the struc-ture. It should be noted, however, that equationsof equi-llbrium at points of the fixed ring KLMNO were not taken in-to account since the rigid fixation is capable of providingany reactionsthat are needed for equilibrium. Similarly,the equationsof equilibriumof the axial forces at themiddle ring R’GHJwere not consideredbecause they werebalanoedby equal and oppositeforces arising from the sym-metric other half of the cylinder,ae mentionedbefore. Theforces and moments arising in the two symmetrichalves ofthe cylinderin the T, R, and N directionsdo not balance oneanotherat the middle ring 3’GHJbut add up, since they areequal in both magnitudeand sense. The conditionsof equi-librium of these forces and moments are consequentlyincludedin the set of equations.

The displacementquantitiesobtainedfollow:\

tA = -0.1035 ~B = -0.4676 ~c = -0.4105

ED= 0.0001

kG= -0.70?1

‘A = 2.0291

u~ = 0.3233

‘c = -0.2924

‘D = -0.1146

‘E = 1.3471

% = -0.3165

UG = -0.4035

‘H = -1.7’210

f E = 0,5735

tH=o

~B = -0.9576

‘c = -1.2685

‘D = 0.5806

‘1?= 0.1872

VG = -0.8034

‘Ii= -0.7?22

8F=-0.9239

kJ = 1.0000

WB = -0,7573

‘c = 0.4316

‘D = -0.2831

# i= -0.6?07

‘J

‘r= -0.17’72

‘G = -().4348

‘H = 0.5665

1.(18)

NACA TN NO. 1014 16.

It may be noted that the unit displacementis 0.001 inch,w and the unit rotation 0.001 radian.

Substitutionof the values obtainedinto the equationscorrespondingto the sixth to ninth rows of the operationstable gives the resultantforces actingupon the constraintsat points Y to J in the axial direction. The foroes in the

4 stringersat these points can be obtainedby multiplyingby-1 the forces oaloulated:

. XF = -’79.9084pounds ‘G = -89.1049 pounds(19)

‘H = -O.1O64 pound XJ = + 170.9411pounds

l!heforces correspondto a bending moment

M = 3077.7 inch-pounds (Boa)

and a tensileforce

P = 1.8214 pounds (20b). Eence the rotation of the plane of ring 3’Gl?Jundertaken

correspondsto the applicationof a considerablebendingmoment and a very small tensileforce. The tensileforcecan be eliminatedby a suitableaxial translationof theplane of ring I’GHJ,

The secondpart of the calculationsconsisted,therefore,of the determinationof the distortionsof the cylinder corre-spondingto an axial displacementof ring FGHJ amounting-0.001 inch. The simultaneouslinear equationswere setin the same manner as before, except that now the values

toup

.—

were used, This new system of equationswas solvedbyDoolittle~smethod. Those acquaintedwith the method willrealize that this second solution involvescomparativelylittle work if use is made of the solution of the first set.

. The results are:

..,

NACA ~1$NO. 1014.

17

t~ = -0.0993i

5D = -0.5’728

t~ = 1.0000

s VA = 2.4041

U3 = 0.4899.

‘c = 0.06’74

‘D = 0.0611

Tz = 0.2400

% = 0.4055.

‘c+= 0.31’76w

‘H = –0.2376

‘J = –0.4981

f~ = -0.5039

k~ = -0.5725

5= = –1.0000

‘B = -0.4650

‘c = –0.8912

‘D= 0.1735

VF = 0.3407

‘G= 0.’7220

~H . -0.0716

t~ = -0.58037

5F = -1 ● 0000

tJ = -1.0000

‘B = -0.7013

‘c = 0.3489

I(21)

‘D = -0.0900

‘I’= –0.2371

= – 0.1459‘G

‘H = 0.2123~

The unit displacementis again 0.001 inch, and the unitrotation0.001 radian.

. Substitutionof the values obtainedinto the equationscorrespondingto the sixth to ninth rows of the operationstable gives the resultantforces actingupon the constraintsat points F to J tn the axial direction. The forces in thestringersat these points can be obtainedby multiplyingby-1 the forces calculated:

‘F = -87.6409 pounds ‘G = -126.0882pounds

(22)‘H = -257.4269 poundss ‘J = -171.9484pounds

NACA TN No, 1014

These forces correspondte a resultantforce

F = -643.1044 pounds (23a)

and a bending moment

1.{ e -1.821 X 10 inch-pounds (23b)

Obviously,the two solutions- namely, those corre-spondingto the pure rotationand the pure translationsrespectively,of ring 3’GHJ- can be combinedin such amanner as to representthe two loading cases of pure bend-ing and pure compression. The solutlonmay be obtainedbysolving in each case two simultaneousequations. Becauseof the great differencein the numericalvalues~ however,it Is quicker and just as accuratefor practicalpurposesto correctfor the effect of tension in the followingmanner in the loading case correspondingto bending:

The tensileforce oausedby the rotation is 1,8214 pounds,

The compressiveforce due to the translationis 643.3044pounds,

A combinationef the rotationundertakenwith a trans-lation of 1.8214/643.1044= 0.00283 units eliminatesthetensile force and introducesan ad~itivebending moment ofless than one-hundredthof a percent of the originalmoment.

The final pattern of distortionsis shwn in figures11 to 14.

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.

In a similarmanner, the loading case correspondingtopure compressioncan be dealt with. The displacementscalcu-lated for pure translationmust he combinedwith those calcu-lated for the pure rotationmultipliedby the factor18.21/3077.7= 0.00593.

The final pattern of distortionscorrespondingto pu”recompressionis shown in figures 15 to 18.

.

HACA TN NO. 1014

Calculationof the Stresses

19

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.

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The averagenormal stress In a segment of a stringerbetween two adjacentrings ts equal to the differencebetween the axial displacementsof the end points of thesegmenttimes Youngfs modulus of the material dividedbythe originallength of the segment. The average siiresswas calculated.from the displacementvalues obtainedasshown in the precedingsection, The values of the stressare plotted against the distance of the stringerfrom thehorizontalaxis of the cylinderin figures19 to 22. Thecurves shown correspondto either a pure bendingmoment of35,000 inch-poundsor to a pure compressionof 1286 pounds.The former value was chosen in order to permit a comparisonwith experimentalresults.

Theory and experimentagree in obtaininga practicallylinear stress distributionover the ma~or portion of boththe completeand the cut sections. The straightlines,however, do not coincide. The reaeon for the discreps.ncyis the differencein the locationof the centroidsof theactual and the simplifiedmonocoque cylinders. Deviationsfrom the straightline occur in the neighborhoodof thecutout. The nature and the magnitude of these deviationsare practicallythe same in experimentand calculation.

The shear stress in a panel dependsupon the displace-ments and rotationsoccuringat all the four cornereof thepanel. If in figure 23 pcint A is displacedaxially througha distance t.A* the shear stress induced in the panel is

fs = -(1/2)(G/L)tA

provided the positive senee of the shearingstress actingin the eheet is as shown in figure 23, It followsfromequation(ha) and the sign conventionadopted that theshear stress caused by a tangentialdisplacementof anamount u~ of pOi12t A is

.

f8 = at(fJ/a)uA (24b)

Similarly the shear stress caused by a unit radial displace-ment VA of point A is

fe = a7(G/a)vA (24c)

NACATNNo.1014 20

Finallya rotationWA oftheringsectionat A givesrisetoa sheeringstress

Displacements

f~ = c@(G/a)wA (2kd)

androtationsat theothercornersofthepsmelcontributesimilerquazrbitiestotheshearstress,butcaremustbe takentousetiepropersi~. Thetotalshearstressis

f* = (1/2)(G/L)(-~A+~B+~c‘~)+~(G/a)(UA+U3-~-uD)

+% (G/a)(-VA+VZB-vc+vD)++(G/@) (w~+”~-wc-W) (24)

Substitutionofthevaluesoftheconstantsandthedisplacementquantitiesyieldsthesheerstress5nanypenel.Figures24end25 containshearstressdistributioncurvesforpurebendingandpurecompression,respectively.A

* ccnrpariscnof thecalculatedvalueswithexperimentalcrnesisnotwellpossiblebecauseof thelimitednuniberofmeasurementsandbecauseof thesimplifyingassun@ionsofthecalculations.E~er@ntal straindataarenotavail-ableforthefullsection.3iIthecutsectiontheaveragemeasuredvalueinthepanelsadjacenttothecutoutwas454PSI [fig.36ofreference6),whilethecalculatedvaluewas40.psi.Thislatter,however,wasobtainedasa smalldifferenceoflargequantitiesandisnotreliableforthisreason.Measurementsinthefullsectionwere

. madeina cylinderhav- thelargecutout.Theresults(seefig.40 ofreference6)indicateconsiderablyhtgherstressesinthepanelsof thefullsectiontheainthose

. ofthecutsectionnearthecutout.Moreover,a chemgeinthesignofthesheeralsowasolserved.

Thecalculationof thebendingmoments,shearforces,endtensileforcesintheringscanbe carriedoutaccord-ingtotheprinciplesstatedinreference4, lhthepresentcaeethemaxtiranbendingnmnentinthefullringat theedge

‘ of thecutoutwasfoundtobe 2.64tnch-poundswhentheload-ingofthecylinderwasahendingmomentof 35,000fich-pounde.Themomentdiagramisshowninfigure26.

..—

.-

NACA TN No. 1014 21

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.

.

8

.

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.

#

The 2.64-inch-poundbendingmement is very i!lsig-nificantas comparedto the applied bendingmoment of 35,000inch-pounds.,Nevertheless,it causeshigh stressesbecauseof the small moment of inertiaof the ring eection, Themaximum stress is 2,64 X 850 = 2240 psi accordingto thetic/I formula,

When the applied loading is a compressiveforce ratherthan a bendingmoment, the moments in the ring are foundto be similar to those just discussed. For this reason themoment diagram is not shown.

DEVELOPMENT03’A STEP-BY-STEPAPPROXIMATIONPROCEDURE

Basic Considerations

The method of calculationof the stress distributionin a monocoquewith a cutoutpresented in the first part ofthis report gave Satisfactoryresults with a reasonableexpenditureof work. It is felt, however,that with morecomplexstructures.-for instance,moneeoqueswith non-syrametriccutoutsor monocoqueswith several cutouts- theoperationstable would become so large that its solutionbythe methods of matrix calculusmight entail too much nu-merical work to be practicable. An effort is made in thispart, therefore,to develop a procedure of step-by-stepapproximationssuitable to cope with these complexproblems.

In order to simplify the presentation.the structureshown in figure 27 is used in place of the actual monocoquecyltnder~ This structurewill be referred to as the smalloylinder. The vertical transverseplaneof symmetryS-S isconsideredhere as fixed in space and the points of thecylinderon the four rings are moved relativeto this fixedsection. However, end rings A-A’and AI-A’ are rigfd and canonly undergo rigid body displacements. Because of thedouble symmetryof structureand loading it sufficestolist only the forces and moments caused at points containedin one-quarterof the cylinder. —.—

In the actual calculationsall four rings were firstrotated as rigid bodies about their horizontaldiameters.The amount and sign of rotationwere defined by the sttpu-la,tionthat the intersectionpoint of ring A-A with stringer1 be displaceda distance of 0.003 inch, that of ring B-Bwith stringer1 a distance of 0.001 inch, both in the

-— ...__

NACA TN No. 1014 22

negative x-direction. The axis of rotationwas the horizontaldiameterof the ring. Rings At-At and BJ-BI were rotatedsymmetrically. This distortionpattern correspondsto thatprevailingin the completecylinder(without the cutout)under the action of a pure bending moment, providedringsA-A and AI-A! also undergo a rigid body translationverti-cally downward. The necessaryamount of translationwasdeterminedfrom the reo,uirementthat at each point alongring B-B the forces and moments caused by the displacementshad to add up to zero force and moment resultants. Inpractice,only one componentforce or moment had to bebalancedout at any singlepoint, after which all the otherpoints were found to be automaticallyin equilibrium.

._As the next step the unbaJ.ancedforces and moments

were calculatedthat arose in the structureof figure 27when the displacementsdeterminedin the precedingpara-graph for the complete oylinderwere applied to the cut cyl-inder of figure 27. Be~ause the loadingdid not involveshear fcrces,unbalanceswere found only in the x-directionand at points along the edges of the cutout.

It was anticipatedthat the distortionpatterh would beinfluencedpaternally by the cutout only in the neighborhoodof the cutout. Consequently,additionaldisplacementswouldhave to be undertaken only at the intersectionpoints ofstringers7, 8, and 9 wtth ring B-3. This restrictionmateriallydecreasedthe amount of work involved in thesolutionof the problem. However, additionalrigid bodytranslationscf the rings were necessary in order tc insurethat the axial forces would add up to a zero resultantacross field B-BI.

The unbalanceswere eliminatedby displacingpoints 7, 8,and 9 in the x-direction. The required displacementswerefound by solvinga 4-by-4 matrix which includedthe equi-librium conditionsin the x-directionat the three points,and the requirementof a zero resultantforce in the x-directicnin a transversesection acrossfield B-Bf. Thesedisplacements,of course, gave rise to unbalanced tangentialand radial forces and to moments in the plane cf ring B-B.The unbalanceswere eliminatedby undertakingsuitabletang-ential and radial displacementsand rotationsat points 7and 8 of ring B-B, and a suitableradial displacementatpoint 9. Because of the symmetrypcint 9 could not-undergoany rotation or tangentialdisplacement, The magnitudesofthe displacement=were determinedby solving a 7-by-7matrix.

NACA TN NO. 1014 23

The displacementsrenuired.for balancingthe forces andmoments in the plane of ring B-B threw back unbalancesintothe x-direction. They were again eliminatedby displacingpoints 7, 8, and 9 in the x-directionand undertakingasuitableamount of rigid body translationof ring B-B inthe x-direction. After this, it was again found necessaryto balance the plane of ring B-B as before. The displace-ments needed for this last balancingcaused insignificantunbalances(about 1/2 of 1 percent) in the x-direction.

After‘allthese displacementswere undertaken,points7, 8, and 9 could be consideredto be in equilibriumforpracticalpurposes. However, a check calculationshowedthat point 6 was considerablyout of balance in the planeof ring B-B. Hence point 6 was now moved and the un-balancesthrown back by these motions upon point 7 werebalancedby moving points 7, 8, and 9 once more in theplane of ring B-B. Again sizableresidualswere found toexist at point 6. The motions of point 6 reauiredforbalancingthe forces and momente at point 6 causedrela-tively large unbalancesat point 7, It was found that theconvergenceof the procedurecould be acceleratedby movingpoints 6 and 7 now simultaneously. After this step all theunbalanceshad values which COUIA be consideredas negligible.The establishmentof a check table, however, showed that the

MAB - liB3,resultantforce quantities ‘AB’ ‘BB!‘ and

ras well as the residualsat point 5, were too large. A fewadditionalmotions sufficedfor reducingthese quantitiestopermissiblevalues.

NumericalCalculationof the Equilibrium

of the Small Cylinder.

.

The completeoperationstable is presented as table 2.Because of its large size it is shown symbolically,and thesymbolsare explainedin table 3. SymbolsP, Q, R, and Shave two values each, one correspondingto the completecylinder(no cutout)and the other to the cylinderwith thecutout. It is noted that the effectivewidth of sheet attachedto stringer8 is reduced when the sheet in the panel betweenstringers8 and 9 is cut out. The entries in the tablewere calculatedaccordingto the principlesdiscussedin thefirst part of this report.

NLCA TN No. 1014 24

.

●

✎

Individualdisplacementsof the points situatedalongring A-A are assumed in spite of the fact that the ring hasto be consideredas a rigid bodY. In the establishmentofthe operationstable symmetricmotions of the entiresystemwere consideredthroughout. Consequently,wheneverthenth point of ring B-B is moved througha unit distance,sayin the tangentialdirection,point n’ of ring B-BQ point nof ring B1-B’, and point nl of ring B’-B’ also are movedsimultaneouslylikewise~ The effect of these simultaneousmotions was duly consideredwhen the operationstable wasestablished. The effect is noticeablein the entriesreferringto stringers2 and 8. The points on ring B-Balso are affected. The unit displacementis 0.001 inchand the unit rotation is 0.001 radian.

Similarly,whenever a point on ring A-A is displaced,the three points symmetricallysituatedto it are also dis-placed. A motion of a point on ring Al-At , however,hasno effectupon the forces listed in table 2. On the otherhand, the displacementof a point on ring A-A and stringers2t and 81 influencesthe entries in rows 2 and 8 in table2, as may be eeen from the numericalvalues given in table3.

!l!able4 lists the forces and moments’caused by rigidbody displacements. A rigid bo&y displacement1s definedas a set of displacementsof a number of points duringwhich the distancesbetween the points do not change.The effeot of a rigid body displacementcan always be oalou-lated as the sum of the effect of the individualdisplace-ments that constitutethe rigid body displacement. In table4 all values are listed for the completecylinder(no cutout).The rows marked ~A~ containthe forces and moments cnusedby a rigid body rotation of ring A-A, the rows marked W3B

thoee caused by a rfgid body rotation of ring B-B, and therows marked ~AA those caused by a vertical downwardtransla-tion of ring BE, The magnitude of these motions are so de-fined that the displacementof point Al is -0.003 inch inthe x-directionfor the rows marked

‘AA and the displace-ment of point B1 is -0.001 inch for the rows marked ‘BB”In both cases the horizontaldiameter of the ring is the axisof rotation. In the downwardtranslationcorrespondingtothe rows marked ‘AA the displacementis 0.001283 inch.It is noted that a vertical downwarddisplacement T ofthe intersectionpoint of a stringerwith a ring defined by

NACA q~ No. 1014 25.

*

.

.

.

an angle q measuredfrom the verticaldirectionas shownin figure 27 can be consi.d.eredas the sum of a tangentialdisplacementof q sinQ and a radial displacementof qcOsQ. The sum total,ofall the rigid body displacementslisted is a group operationwhioh gives vanishingresidualforces and moments throughoutexcept for the forces actingen ring .4-Ain the x-direction. The latter add up to apure coumle of 49,000 inch-poundsabout a horizontaltrans-verse axis.

In some instancesconsiderationof all the displace–ments simultaneouslyreduces the work of computation. Torinstance,in the case of a rigid body displacementof a ringIn the x–directionit is self–evidentthat no shear w*11arise in the panels. The only forces caused by such adisplacementare due to the shorteningof the stringers.

The displacementscontainedin table 4 constitutethesolutionof the problem of bending of the cylindernot havinga cutout. Since it iS anticipatedthat the effect of thecuto~twill be restricte~to the immediateneighborhoodofthe cutout, in the balancingprocedure to follow only aportion of the operationstable will be used. This por-tion is presentedas table 5. It containsonly the displace-ments of points 7, 9, and 9 and four ri,gidbody translationsof the rings, all related to the cylinderwith the caztout.

The uppe~left 10-by–10corner of the operationstableis identicalwith the correspondingportion of table 2;

ML

HPQ

Its.

The eleventhrow representsa rigid body translationin the x–directionof ring A-A. The eleventh column containsthe contributionsof the individualand group displacements

. to the resultantaxial force acting in a completetransversesection of field A—B. Similarly,row 12 c5rrespontls’to arigid body x-translationof ring B—B (comhinettwith a

. simultaneoussymmetrictranslationof ring Bt-Bl), and column12 containsthe resultantx–forces in a transversesection offield B-B~, Row 13 containsthe forces and moments caused bya rigid body rotation of ring B-B about its horizontal

NACATN No. 1014 26

dianetersuch that t -0.001 inch for the intersectionpoint of stringer1 wi;h ring 3-3. Simultaneously,ofcourse,ring B1-B~ 5s also rotated symmetrically. Incolumn 1% are listed tihecontributionsof the individualand group operationsto the expression (M~~ - MBB!)/r

—where ‘AB is the moment about the horizontaldiameterofthe cylinder in a transverseeection through field A-B,Ii*B the correspondingquantity in field 3-3~, and r isthe radius of the cylinder. Finally the last row representsa rigid body vertical downwarddisplacementof ring AA.(Of course ring A~-A1 must also be displacedsimultaneously,but this displacementwill not have any effect upon thequantitieslisted in the operationstable.) The magnitudeof the translationis such that the radial displacementvof the intersectionpoint of stringer3 with ring A-A is+0.001 inch. The last column containsthe contributionsof the individualand group displacementsto the verticalshear force resultantacting on ring A-A.

..-

On the assumptionthat the displacementscalculatedintable 4 for the complete cylinderrepresentin first approxi-mation the displacementsof the cylinderwith the cutout$ “---the values are substitutedin the operationstable of the ..cylinderwith the cutout. !Cheonly unbalances corre-spondingto these displacementsoccur at points 38 and B9in the x-direction,becausethe displacementpattern oftable 4 does not aontain shearing deformationsin fieldBB!. The unbalancesare

‘B8 = 71.17 pounds ‘B9 = 613.32 pounds (25)

O.orrespondingto these values the sum of the axial forcesin a section through field B-B! is not zero but

. ‘3Bt= 755.67 pounds (26)

and the differencein the moments in the sections throughfieldsAB and BB1 is (divided,by the radius of the cylinder)

.(MAB - 1+13B,)/r = 744.8 inch-pounds (27)

.Points B7, 3.8,and 139are now balanced,and the a~fal

force resultantin field B-B? is reduced to zero by suitablex-displacementsof points 37, B8, and B9, and by suitable

NACA TN NO. 1014 27

rigid body translationsof r$ngs BB and AA In the -direction.The upper left 3-by-3 cmner of the mat?ix shown in table6 containsthe axial ferees eauso~ by the axial dtsplaue-ments of tho three psiato il%question- The fourth rowgives the axialforcesuaustd by ILunit axial rigid bo&yx-translationOf %ixig B-1) sambined with an axial trans-lation of ring A-A throughu distunoe of three units inthe x-direction. Rtng LA has to ‘bemoved tn order to in-sure that points 31 te S6 h,umet thrown aut of balance be-aause of the rigid ‘b~d?~S~~lat$on of ring -B. The com-bined operationthus ~~fiae~ As demoted as g group. Itmay be seen that all the figars~tn table6 either are taken -direetl~from table5* w are eQmbSaationsof values listedin table 5,

In order to balance ths X-rgsidualsobtained,the setof equationsrepresente~by the matrix of table 6 anctthe~ight-handside members gtve~ in oquat40ns(25) and (26)are solved by the matsix method. The results are

t,.= 0.0253 ~B = 0.12816 ~9 = 1c9190 ~group= 0.0’7 (28)

Substitutionof the above displacementvalues into partof table 5 gives the residualforces and moments sating inthe plane of ring B-B at the location of stringers7, 8,and 90 (The residuals tn the x-directionwere balancedout in the preaedingstep of tha calculations,) The newresidualsare:

T37= -,0,480 pound R

B9= 3.352 pounds

~B8 = -21.868 pounds ‘B7 = -0.03717 inch-pound

. (29)’137= -0.07256 pound ‘Be = -0.S487 inch-pound

‘B8 = -1.5798 poun~.

To eliminatethe unbalanceslisted tn equations(29) theseven equationsrepresentedby the matrix of table 7 (which.is just anotherpart of table 5] to ether with the right-hand members given in equatione(29f are solved by thematrix method. The solution1s;

NACA TN No. 1014.

-28

‘B7 = 0.079 ‘B? =-0.859

UB8 = -1.07’9 vB8 = -2,602

‘B7 = -1.429

WB8 = 1.904 (30)

‘B9 = 6.68

When these displacementsand rotationsare undertaken,theforces and moments acting in the plane of ring B-B are inequilibriumat points B7, B8, and 39. However, the equi-librium of the forces acting at these points in the =direc-tion has been disturbed. The unbalancesthrown back in thexdirection are calculatedagain from table 5. They are

‘w =-11.06 pounds %8 = -10.85 pounds ’39 = 41.18’7pounds

(31)

No unbalancedaxial foroe results in a transversesectionof field BB1. The residualsare small as comparedto theoriginalones. Nevertheless,they are eliminatedby usingonce more the matrix @f table 6. The necessarydisplacementsare

tB8=-0.o09684

%9 = 0.1232 1 (32)t = 0.00274group

,)After these axial displacementswere undertaken,theunbalancesin the plane of the ring are:

‘B7 = 0,222 ‘B7 = -0.006 ‘B7 = 0.0188

’38 = -1.604 ‘B8 = -0.062 ~B8 = -0.054 (33).

‘B9 = 0.371;

. Use of the matrix of table 7 gives the followingdisplace-ments and rotations:

.‘B? = 0.010 ‘B7 = -0.0637 ‘B7 = -0.121

—

‘B8 = -0,0844 ‘3!3= -0.213 ‘B8 = 0.144 (34)

‘B9 = 0.553

..

NACA TN NO. 1014 29.

The axial unbalancescaused by these distortionsare negli-% gibly small.

Substitutionof all the preceding displacementvaluesinto the operationstable (table 5) reveals that a differ-ence now existsbetween the moments transmittedthroughfieldsA–B and B-BI:

,.(~AB – MBBI]/r = -28,78 inch-poundeper inch (35)

.

This can be eliminatedby rotatingring &B through an angle(in 1/10,000rad)

.

‘BB = -0.C05 (36)

The unbalances caused at the differentpoints by this rotationare found to be negligiblysmall.

Points B7, B8, and B9 can now be consideredas completelybalanced. Substitutionof the displacementvalues corre–spendingto all the individualdisplacementsof point B7 in—to that portion *f table 5 which representsthe interlinkti~ebetweenpoints B6 and B7 shows that point B6 is out of balance.The unbalancessre\

‘B6 = 2.082 pounds(37)

‘B6 = 3.01 pounds RB6 = 1.123 pounds‘B6 = 0.893 inch-pound

The residualforces and moment listed in eQuations(3’7),togetherwith the matrix of table 8, constitutea system offour linear equationswhich permits the calculationof the fourdisplacementsof ~otnt B6 nece9sarY to balance out the r~=idualq..The Uis~lacementsare

t = 0.002582B6

. UB6 = -0.01287 ‘B6 = 0.20924 ‘B6 = 0,28929 (3’8)

These displacements,while balancingpoint B6, throw unbalancesupon points B5 and B7. The former were not recorded:thelatter follow:

NACA TN NO. 1014 30

.

TB7 = 4.1~68 pounds ’37 = -0.9831 pound.

}

(39)

‘B7 = -0.0826 pOUld ‘B7 = 0.3509 inch-pound

The three quantitiespertainingto the plane of the ring arenow balanced out using the matrix of table 7. The displace-ments obtainedare:

.

.

.

.

.

‘B7 = 0.03607 ‘B7 = -0.23349 ‘B? = -0.11682

%8 = -0.02859 = 0.01284 1‘B8 ’38

= 0.07393 .(40)

PB9 = 0,080~2J

The unbalances caused at B6 by the above–listeddisplacementsof B7 are:

T = 3.778 poundsB6

R = 0,812 poundB6

}

(41)

‘B6 = 0,601 pound. ‘B6 = 0.522 inch-pound

‘J!heseresidualsare again eliminatedwith the aid of tablea. The followingdisplacementsare obtained:

-—

~B~ = 0.00092

}

(42)UB6 = 0.02233 TB6 = -0.15102 ‘B6 = 0.03928

The effect of these motions on point 7 is found to be:

T! = 3.33B7

x=B?--oc393~

1

(43)R= -0.6636 N = 0.402B7 B6

Comparisonwith the values shown in equation(39) fn-dicates that this process is very slowly convergent,if at .—all. A rapid eliminationof the residualsat %oth points6 and 7 can be had only hy moving both these points at thesame time. The motions Udertaken are:

NAGA TN NO. 1014 31.

‘B6 = 0.012 = 0.019‘B7.

= 0.11 -0.17‘1

(44)‘B6 ‘B7 =

= 0.03‘B6= 0.08

‘B7 J

. At this stage of the relaxationsall the remainingunbalanceson points 6, 7, 8, and 9 are consideredas negligiblysmall.A check table is set up and 1s presentedas table 9. It. indicatesthat point 5 is out of balance.

The check table also shows unbalancesfor XAB, XBBf,

and (MAB - MBB,)/r. These residualsprobably are due to.,

some errors in the numericalcalculations, All the residualsare reduced to negligiblysmall quantitiesby additionaloperationscontainedin table 9.

. Figures 28, 29, and 30 contain the axial stressrthebending stress, and the shear stress distributions,respec-tively, in the small cylinderas calculatedfrom all the

. displacementsdeterminedin this section.

NUMERICALCALCULATIONOF THE EQUILIBRIUM

OF THE LARGE CYLINDER

It was hoped that applicationof the procedurejustshown would result in establishingthe equilibriumof thelarge cylinder in a reasonablenumber of steps. This an-ticipation,however, was not fulfilledand the calculationsbecame so time consumingthat they cannot be recommended

, for routine work, although the results obtainedwere ingood agreementwfth experiment. .— —

The system of designatingthe individualpoints is shownin figure S1.. As may be seen, the ‘tlarge’icylinderis’justa non-simplifiedversion of the same cylinderthat was calc”u-”-lated by the matrix method in the firet part of this report.Only one-quarterof it need be consideredbecause of thesymmetry. This quarter containsone rigid and three non-

“rigid half-rings(one of the latter is cut) with altogether35 points which have a total of 97 degrees of freedom ofmotion.

HACA TN NO. 1014 32

.

The operationstable is representedsymbolicallyintable 10. The squaresdenoted bY 1, 2, and 3 are identical.with the squaresthat are similarlysituated in table 2.Table 11 containsthe square designated4, and table 12those denoted by 5, 6, 7, and 8. SymbolsA to Z in thesetableshave the same meanings as before (see table 3), andthe symbols a to u are explainedin table 13. .—

. The principleused in solving this operationstable wasthe same as that discussedIn connectionwith the small cyl-1inder. First rigid body displacementswere undertakenwith

. all the eight rings in order to find the solutionfor thecompletecylinder(no cutout). This involvedrotationsofrings D-D, C-C, B-B, and A—A in the ratios 1:3:5:7,andverticaldownward translationsof the last three in theratios 1:3.02:6.01. Next the displacementsobtainedweresubstitutedin the operationstable for the cylinderwiththe cutout and the unbalanceswere calculated. These werethen balanced out by solving the matrix of all the x–dis–placementsconsideringonly points 6 to 9 on rings B-B andO-C, and 5 to 8 on ring D–D. The displacementsundertakencausedunbalancesto arise in the planes of the three rings..Three matrices were set Up to take care of the forces andmoments in the plane of each ring individually. The un–

. balances in the plane of one of the rings were eliminatedfirst by solvingthe correspondingmatrix. The unbalancescausedby these displacementsin the plane of the next ring,togetherwith the originalunbalancesthere, were thenbalancedby solvingthe correspondingmatrix, and so on.Because of the unexpectedlystrong interactionbetween the .rings the matriceshad to be solved many times before theunbalancesvere reduced in all the rings simultaneously.The displacementsundertaken in the planes of the ringsduring this balancingprocedurethree unbalancesback Inthe x-directionwhich necessitateda repetitionof theentire‘procedure.

.

.

At the beginning,the unbalances in the x–directionde–creasedafter each completebalancing in the plane of therings but later they began to increasegradually. .

At thesame time, the displacementsin the neighborhoodof thecutout increasedsteadilyand tended to attatn unexpectedlylargevalues. Consequently,the procedureadopted was foundto be divergent. It is possible,however, that the divergencewas caused eitherpartially or wholly by a slight error-“inthe operationstable. .

NACA TN NO. 1014.

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It was also observedthat point D8,underwent large out-ward displacements(in the negative r direstion) while inthe solution of the ~implifiedstructurein the first partof this report the dis~lacementof the correspondingpointwas small and inwar~. Because of this the displacementpattern of the rings was arbitrarilychangedto conformbetterwith that foun~ in the case of the simplifiedstructure, The procedure of balancingwas then continuedas before and was feund te converge,though S1OWIY. Itmight be mentionedthat in the tests described in reference6 both inward and outward deflectionswere observed.

At suitableetages of the procedureagain some additionalpoints (B5, C5, C4, and D4) had to be displaced in order toreduce the’residualsall over the etructureto negligiblysmall quantities. The final displacementsobtainedarelisted in table 14, and the final reeldualforces and momentsin table 15. The stresses calculatedfrom the displacementsare shown in figures 19, 20, 24, and 26.

It may be seen from figures 19 and 20 that the axialstress distributionis very much the same in the solutionscorrespondingto the simplifiedcylinderand the largecylinder. The straight-lineportions of the diagramsarepracticallyparallel althoughnot coincident. The reasonfor the shift is that the location of the centroid of thecross section of the simplifiedcylinder is not the sameas that of the large, and consequentlyalse the actual,cylinder. It can be anticipated,therefore,that the agree-ment is better between the experimentalcurves and thetheoreticalcurves calculatedfor the large cylinderthanbetween the experimentalcurves and the theoreticalcurvescalculatedfor the simplifiedcylinder. This is borne outby figures 19 and 20. Altogetherthe agreementbetweentheory and experimentis good.

In the shear curves of figure 24 and the bendingmomentcurves of figure 26 the agreement is good between valuescalculatedfor the completeportion of the cylinderon thebasis of the simplifiedand the large cylinders. Consider-able deviationseccuy in the cut field. This could be ex-pected since the simplifyingassumptionschanged the mechani-cal conditionsin this,region.

.liACATN NO. 1014 34

A SIMPLIFIEDAPPROXIMATESOLUTION

.

.

The pro%lem can be simplifiedradicallyby assumingthat the rings are infinitelyrigid in their planes and byestablishingthe equilibriumof the x-forces only. Thispropositionwas worked out on the basis of the operationstable of table 1. The elementsrelated to the equilibriumof t’heaxial forces are containedin the upper-left9—by—9corner of the operationstable.

The equationscorrespondingto the assumed rotationof the rigid end ring were solved by the matrix method.The displacementsof points A, B, O, D, and E were found tobe --.-..— .—

tA = –0.1482 ~B = -0.4340 Ec = -0.3799

)(45)

t~ = 0.5488 JThe axial stresseswere calculatedfrom these displace-

ments. They are shown in figures 32 and 33 togetherwiththe curves of axial stress calculatedin the first part ofthis report. The agreementwas found to he excellentbetweenthe present approximatesolutionand the exact solutionofthe problem of the simplifiedcylinder. Of course,theapproximatesolution does not give any useful data for thecalculationof the bending moments in the rings and the shearstressesin the sheet covering.

CONCLUSIONS

The stress distributioncau~ed by a pure bendingmomentin a cylindricalreinforcedmonocoque cylinderhaving asymmetriccutoutwas investigatedby severalmethods of calcu-lation. The results were comparedwith data obtainedin ex-perimentsdescribed in reference6. The main conclusionsfollow?

1. The operationstable of the problem as defined in theSouthwellmethod can be set Up easily if use IS made of theformulascontainedin figures ‘7to 10. .- .—

NACA TN NO. 1014 35

—

2. The set of simultaneouslinear equationsrepresentedby the operationstable can be solved by the matrix methodas shown in reference4 if the number of unknowns is nottoo great. (The 30-by-30matrix shown in table 1 can besolved in from 1 to 4 days dependingupon the operatorandthe calculatingmachine..)

3. The calculatedaxial stressesare in good agreementwith the experimentaldata presented in reference6. (Seefigs. 19 and 20.) The tliffereneein the Ioeation of theneutral axes correspondingte test and calculation(thelatter labeled simplifiedcylinder)is due to the fact thatthe location of the centroidof the actual structurediffersfrom that of the simplifiedstructure.

4. The step-by-stepproceduredevelopedin the secondpart for solvingthe operationstable of the so-calledZargecylinder(table 10) was slowlyconvergentand is not recom-mended in its present form for practicaluse, The resultsobtainedby it for the axial stressesare in good agreementwith test results. (See figs. 19 and 20,)

5. Experimentalvalues for the shear stress in theoylindershown in fi~ure 1 were availableonly in the cutsection. They de no;values..However, thereliableas stated in

PolytechnicInstituteBrooklyn,N. Y,,

comparefavorablywith the calculatedexperimentalvalues are not consideredreference6.

of Breoklyn~July 1945.

.

NACA Th NO. 1014

REFERENCES

36

.

1. Southwell,R. V.: RelaxationMethods in EngineeringScience,A Treatise on ApproximateComputation.ClarendonPress (Oxford),1940.

2. Hoff, N. J., Levy, Robert S., and Kempner,Joseph:NumericalProceduresfor the Calculationof theStresses in Monocoques. I - Diffusion of TensileStringer Loads in ReinforcedPanels. NACA TN No. 934,1944.

3. Heff, N. J., and Kempner,Joseph: NumericalProceduresfor the Calculationof the Stresses in Monocoques,II - Diffusion of Tensile Stringer Loads in Rein–forced Flat Panels with Cutouts. NACA TN NO. 950,1944.

4. Hoff, N. J., Libby, Paul A,, and Klein, Bertram: Nu-mericalProceduresfor the Calculationof the Stressesin Monocoques. 111 - Calculationof the BendingMo-ments in FuselageFrames. NACA TN No, 998, 1946.

5. Hoff, N. J., Klein, Bertram,and Libby, Paul A.: Nu-merical Proceduresfor the Calculationof the Stressesin Monocoques. IV - InfluenaeCoefficientsof CurvedBars for Distortions in their own Plane. NACA TN No.999, 1946.

6. Hcff, N. J., and Boley, Bruno A.: Stresses in and GeneralInstabilityof Monocoque Cylinderswith Cutouts. I –ExperimentalInvestigationof Cylinderswith a Sym–metric Cutout Subjectedto Pure Bending. NACA TN NO.1013, 1946.

.—

7, Doolittle,M. H.: Method Employed in the Solution ofNormal Equationsand the Adjustmentof a Triangulum.u. s. Coast and GeodeticSurvey Rep, 19’7E,pD. 115–120.

8. Dwyer, Paul S.: Doolfttle ‘Technioue,Annals of M~thematicalStatistics. vol. XII, no. 4, Dec. 1941, pp. 449-458.

.

.

NACA TN No. 1014.3?

API?ENDIX

To show how figures 7 to 10a representingthe four panelproblemmay be put tieuse In settingup the operationstable,two numerical examplesare worked out. First, use is madeof figure 7 to determinethe forces and moments introduced

. at the constraintswhen point B of the sitiplifiedstruc$ur@is displacedaxially througha positiveunit distance8$0.001 inch. The followingdata are needed in t.haf~teu-

. lations: —

E = 10.3 x 10° psistr

AS%rny =A = 0,18Q7 ia.aI!tE~H

aIl,N = 12.86 in.I

%,IZ = %I,IV = 3.927’in. I’11 = ‘III = ‘IV = 0“385 ‘ste’=.3.97-X 106pi

IG1=O I(@t)&= 47.64 lb. J

The values of the coefficients m were calculatedfrom. the simplifiedformulas suggestedin the conclusionof refer-ence 5. They were checkedby the values taken from figures86 to 93 of reference5 for the smallestand largestvaluesof ~ and Y, respectively. All calculationswere oarried

. out by slide rule. The results are:

.

N&CA TN HO, 1014

‘%)1,11 = ‘~t)II~,~v= ‘0.499

(~=)~,#~ = (~r)~I~,~V= 0.039

1

(lb)

(t2n)I,II = (~n)III,IV = ‘0-0033

For A the forces and moments actingupon the constraintsthen become,accordingto figures 7 and 10a:

X* = (47.64](32.86)/(4)(3.927)= 38.9 lb1

‘A=-(0.499/2} 4’7.64 = -11.9 lb

1

(2a)

‘A = -(cL039/2) 47.64 = –0.934 lb

‘A=-(0.0033/2)(47.64)(3.927)= –0.305 in.-1$

At B the motion causes:

‘B = -10.3 X 103(0.1877)[(1/12.86)+(1/9.64)]

+(47.64)[2(12.86)+9.64-T1/(4)(3.927)]= -433 lb

= -(0.499/2)4?.64

1“

(2b)‘B

= -11.9 lb

‘B = (0039/2) 47.64 = 0.934 lb

NB = -(0.0033/2)(47.64)(3,927)= -0.305 in.-lb J. At point C there results:

‘c = (47.64)(12.86+ 9.64)/(4)(3.927)- 68,1 lb

‘o=RG=NC=O

. (2C)

.

NACA TN No. 1014 39.

For F and G the values are:.

XF = [10.3 X 103(0.1877)/(9.64)]- (47.64)(9.64)[1/(4)(3.927)]

= 146.8 lb.

x = (47.64)(9.64)/(4)(3.927)= 29.2 lbG

‘F = ‘G = 11.9 lb

‘F = -RG = 0.934 lb (2d)

‘1?=RG= 0.305 in.-lb

To illustratefurther the use of figures 7 to 10a, theeffect of a tangentialdisplacementof point G throughaunit distance of 0.001 in. is investigatedwith the aid offigure 8.

. In the courseof findingthe ring influencecoefficientsneeded in the calculation, the ratios

.

(EI)R/L,(EI)R/L2, (EI)R/L3

have to be determined. The moment of inertiaof the ring

plus its effectivesheet is found to be 8.05 X 10–F in.4;L1,~~ is 3.927 in. for arc I’G;and L1ll,IP = 7.854 in. for

ring segment GH..

Further, convenientvalues of the parameters Y and Ehave to be assumed before use can be made of figures 14 to31 for the movable end Influencecoefficientsand figurese50 to 6’7for the fixed end influencecoefficients,or tablesIII and IV, all of reference5. The values Y = 10,OOO ande = 0.25 were found to be the closest choice for the givenring elements. For FG,which subtendsan angle of 22.5°,. the graphs were used; for GH the necessarydata were takenfrom the tables. The final results are: —

.

liACATN NO, 1014.

40

Eor arc 3’G:.

n%~ = 1.539

.

.

tt~ = 41

3’orarc GH:

nGiM = 0.874

nnrx = -0,423

nntM = 0.81

nrrM = 0.281 .

tGM = -0,62’75

t7Jf= 1.538.

n

nn~ = 0.2853

n?~ = -0.792

n%F = 5.54

rGr = 1.461

t;F = -8,10

t%F = 41.4

-nn3’= 0.2562

nnrr = -0.2?3

nnty = 0.871

nrrF = 0.2445

t;~ = -0.641

t;r = 1.530

(3a)

(3b)

These values correspondto a unit displacementof 0.001in. or a unit rotation of 0.001 rad..

When point G is moved, ‘I and Gxv must be set equal

to zero since panels I and IV accordingto the notation offigure 8 of this report are cut out. The parametersmtt art an for arc FG are identicalwith those listed in

. equation(lb). Those for arc GH were derived in analo-gous manner and are:

NACA TN No. 1014 41

.

.

CLtIII,IV=

-0.490

arIII,IV = 0,079

‘-0,0065%I1l,lV = 1(3C)

. The remaining geometricand mechanicalpropertiesoccurringin the calculationsare Identicalwith those given byequation(la).

.The forces and moments arising at the constraintswhen

point G is moved tangentiallythrough the unit distance cannow be written down.

X3 = (0.499/2)47.64 = 11.9 lb1

‘B = (0.499)=(47.64)(3.92?)/9.64 = 4.85 lb

}

‘(4a)

‘B = (0.039)(0:499)(47.64)(3.927)/9.64 = 0.378 lb

NB = (OO038)(Oi499)(47.64)(3.927)=/9~64 Ia 04126 in.-lb,

‘c = -(47.64)(0,499- 0.490)/2’ = -0.1 lb >

‘o = 47.64[(0.499)a(3.927)+(0.490)a(7.854)]/9.64= 14.16 lb

RC = 47.64[(0.490)(0.079)(7.854)-(0.499)(0.039)(3,927)]/9.64

I

(4b)= 1.126 lb

.

‘c J= 47.64[0.490)(0.0065)(7.854)a+(0.499)(0.0033)(3.927)]/9.64

= 1.099 in.-lb.

.

NLCA TN No. 1014

%= -(0.490/2) 4’7.64. = -11.8 11 I

TD = (0.490)z(47.64)(7.854)/9.64 = 9.31 lb

}(4C)

‘D = –(G079)(0.490)(47.64)(7.854)/9.64 = –1.504 lb

‘D = (Q0065)(&490) (47.64)(7.854)2/9.64J

= 0.973 in.-lb’

—x* =xB= 11.9 lb”

% = 41.4 - TX = 36.55 lb

1

(4d)‘r = 8.10 - RB = 7.722 lb

NF = 6.54- NB = 5.414 in.-lb

‘G = -41 –1”538-Tc = -56.698lb

I

(4e)‘G = ‘7.95–0.6275- Rc = 6,1965lb

‘G =-5.38 –o*al - EC= -7.289in.-lbJ —

‘H = ‘O = ‘11”8 lb

‘H = 1.530 – TD = -7.78 lb

I

(4f)‘H = –().641-RD = 0.863 lb

‘H = 0.871 –ND = -0.102 in.-l% J

NACATN No. 1014 43

.

.

.

.

.

.

. .—

TABLEI. OPERATIONSTABLEFOR THE SIMPLIFIEDCYLINDER. .,

.—

44 NAOATN No. 1014--

.TableII. CompleteOPSrationsTablefor&all Oylinder

.

StringerNo. RingAA RingBB

1

.2

3

4

RingAA5

6

7

8

9

1

2

3

4

RiJw13B5

6

7

8

9.

123456789

AC

BEI)

FBI)

FED

FED

FED

ytxWI

FED

F~G

CIA

ytxZt

viT

123456789

TV .Uxw

Yxw

Yxw

Yxw

Yxw

YXP?

Yxz

VIT

HK

JUL

NML

NML

NML

NML

NML

NPQ

RS

— —

. ..——.

.

.

.

.

.

.

“1,. v b.

>

TableXII. sectionsof@rationsTablBforSncll●@linder

‘x T R

SactionA

~ :;44-; -1.872v. -5.39

SectionB

~ 37,82 1.872u -23.1 23.94v 1.872 2$6w -0;S8 ,

P.Sectionc

●

N x T R

SectionF

3 IEI.91 IL55 0.936U-11.5553.45U-97v 0.936-IL.97-2.28w -0.29 8.09 l.y)

337.82 lo&2

Ii XT R N

SectionI

0.29 I 18.9111.55 -0.936 0.298.09 U -11.5553.45.-~97 8.09-1.34)v -0.936U.97 -2.280.63 W -0.29 8.09.-1.X ::$

SectionU . P(nocutout)

u 23.1 -23.94 - o -139.9 0II -16.63V 1.872 -&s6 v 1.872 “5639W 0.58 4.60 w o -:.63 -3:86

~ 18.91-1.lB55 0.9$ -0029 SecticmH -.S(nocutout) sectionMv 0.936 12*97 -2.28 1.30

; ;~%$9z 1.&j2 t 18.91-U*55 -0.936 $.:SectionCl -5.39 u n-*55 53*45J1.97

v -0.936-11.97-2.28 -1:%T 18.91+ti.55 ‘0,9360.29 SectionJ R 0.29 8.09 1.30 0.63v 0.9% -11*97 -2.28-1.30

: ;;:? -1.&72SectionII

SectionQ (nocutti)23.94 ,

v -l.&F! ;4i# /:3’7j:: :.;; ;::5

-1.8720.934429 m 0.58 . -23.94.11.978.09 v 4.6%! -4.56

v 0.;3611:97 -2.2$ L30 sectionK w 4.58 2.60w 0.29 8.09 -1.~ 0.63. ~ 18,91+1.I..55-0.9?4+0.29 SectionS (cutWt)

SectionE . v -0.936-11.w+2.28-3-930: ~M&5 1.872

: -3~”5& q -1.872 0 SectionR (nocutout) -5.39-1.6.63

v -1.872 0 “ -5:39 0 y 18.91-11.55-0.9364*29-16.63 -3.865y -0.936-n.~ -2.28-1.30

●

w o 0

TableIII.SectionsofOperationsTableforSmallCylinder(cont~d.)

XT R N

SectionR (cutout)

“F 18.91-XL.55a.936-0s29V -0.936-11.97-2.28-1.30

SectionP (cutout)

~ -880.920 1.872u o -139.90 -;.63v 1.872 0 -5639w o -1.6.63 .3:%

SectionQ (cutout)

,? 37.82 -1.872u -23.1 -23.94v -l.&p -4.56W -0.58, -2.59

SectionT

? 268.9 -M72v 1.672 0.0925

SectionU

“~ 37.82 -1.872u --23.1 1.142v 1.872 -0.0925w =4.58 0.028

SectionV

‘ ~ 18.91IL55 4.9360.29v 0.9360.571-0.04620.014

,, 1

I.....,,.

x T R N

SectionVl

; lA.Q1—.,- _~l~~+.9X ~m~$lV 0.936-0.571-0.0462-O.OU

SectionW

3 18.91IL55 4.9360.29u 11.557.05 -0.5710:177v 0.9360.571-o.04620.o14w 0.290.177a014 0.0044

SectionX

~ 268.9 0 -1.872 0u o I.&l o 09354V 1.872.0 0.09250w o 0.354 0 0.0088

SectionY

~ 18.91-33..55-0.9364.29u -11.557:05 0.5710.177v 0.936-o.571-ooo462-ooaf&w -0.290.1770.01.40.0044

SectionZ

~ 37.82u -23.1v 1.872w -0.58

.

-1.872M.&+.09250.028

*

x

~ z.&2u 23.1V -1.872W 0.58

\ 18.91v -0.936

; 18.91v -0.936

~ 1.8.91u -IL*55v -0.936w ao29

: p;;

v -0.936w 0.29

3 379=u -23.1v -1.872w -0.5$

f

T R N

SectionU{

1.8721*U2-0.092s0.028

SectionV~

-IL550.936 -0.290.571-0.0462O.OIJ+

SectionVi

u.55 0.9360.29-0.5714.0462-0.014

SectionW

-11.550.9j6-0.297.05-0.5710.177

0.571-o.04620.o140.177-0.01.40.0044

SectionYt

IL.550.9360.297.050.5710.177-0.57M.0462aou(3.177O.ou 0JXL44

SectionZ!

1s872-1J424.0925‘-0.028

● ✌ 4, u , I h .

●

TableIV. SolutionofSmallCylinderWithoutCutout %~

RingAA H~RigidBody Al A2 A3 Ah A5 A6 A7 AtiMotions

A9 ~●

.

X Forces

3 928.8 858.3 656.7‘+

35595 0 -355.5 -656.7-858.3-928.81 -202.8-280>7-21&8 ..-116.30 IJ.6.3”2L$.8 280.7 303.81.2&~-IL52 -10.65-8.15 -’-4.4.00 WJo 8.15 10.65 sL52

613.48566.95433.75234.8 0 -23.4.8-$433.75-566.95-613.48

3 0, 10.U.1.8.7524.51 26.52 24*5L 18.75loll 0 I

1 0 3*37 6.25 8.17 $.84 8.17 . 6.25 3*37 o1.283 0 -13.54-25.02-32.67-35,36-32.67 -25.02-13.54 0

0 -00(% -0.02 0.01 0.00 0.01 -0.02-0.06’ 0

I

I

I

R Forces I

3 0.43 0.40 ?.30 0.16 d -0.16 -0.YI -0.40 -0.431 O.11+ 0.13 0.10 0.05 0 -0.05 -0.10 -0.13 4.JJ$1.283 -0.57 -0.53-0.I$L-0.22 0“ 0.22 o.#!J,0.53 0.57

0.00 0.00 -0.03.-0.01 0 0.01 0.01 O.m 0.00

Moments

3 0 0.26 0.47 0.62 0.67 0.62 0.47 0.26 01 0 0.08 0.I.6 O.zl 0.22 o.2t O*I.6 0.08 01.283 0 -0.34a63 -0.82 -0.89 -0.82 -0.63 -o.% o

0 O.al O.cm 0.01 0.00 0.01 O.co 0.00 0 %’

,1

-.

— R&i BodyMotions

‘AA= 3,.*.1

qM. 1.283

lG--

X ties

-9U..4 w.1 -644.4 -348.9 0 3@.9 644.4 842.1 911.4922.98 852.72 6S2.6S 3s3.2‘ O -3S3.2 -652.65-8S2.72-922.98-IL.53 -10.65 -8.15 “4.40 o 4*4O 8.15 I&65 11.530.05 -0.03 0.lo -o.1o 0 0.1o -o.lo 0.03 4.05

‘FFckes

o -1OXL -18.75 -24.514%.52 -24,51 -UI.75 -10.IL o0 -3*37 -6.2s -8.17 -8.84 -86X7, -6.25 -3*Y7 o0 13.54 25.02 32.67 35.36 32.67’ 25.02 13.54 0o“ 0.06 0.02 -0.01 O*OO -0.01 0.02 0.06 0

R Farces-0.43 -0.40 -0.30 -0.I.6 o 0.I.6 0.30 o.&$o 043-o.llk-0.13 4.10 -0.05 0 4.05 0.lo 0.13 O.u0.57 0.53 O.kl 0.22 0 “0.22 -Ml -0.53 -0.570.00 0.00 0.01 0.01 0 -0.01 -o.ol 0.00 0.00 g

PMamnts H

%o -0.24 -0.47 ,-0.62-0.67 4.62 -0.47 -0.26 0 =o -0.0!3 -0.I.6 -o.21 -0.22 -0.?3, 4.16 -o.mj

o0 .

0 0.34 0.63 0.82 0.89 0.82 O.@ O.% O P

o O.a) O.al -0.01 O.(XI -0.01 O.al O*KI o e*

. . .

t1 8 . ●

, u. s.,

1

,

eS&’bH

Tabla V. Portionof OperationsTablefor8mll CylinderwithCutout %Szo

Forcesinlb.andHands in in.lb. .

P

Motions(%3 s

in (1.001

:’:. % “7 ~% %8 %8% ‘B8’B9” ~xAB$YY-%B~ %*’

yw -957.92 0 1=872o J.8.9111.55-0.9360.29 0 0 613.4-KU().8-1301.2 1.2.w

%7°-139.9 0 -%.63 -11.5553.45-U..978.09 0 0 0 0 0 38.98

v~ 1.872 0 -5.39 0 4.936 1.1.97-2*281.30 0 0 0 0 0 -0.6288.C9A30 0.63 O 0 0 0 0 0.978

~; 3.8:91:::; -0:9362::29 0 1.872 0 27.82-1.872613.4-1072-1557.816.00

u~ 11.55 53.45II-.978-09 0 -139.9 0 -16.63-22.1-23.94 0 0 0 Z1.lo

vBg -0.936 -1.l.w-2.28-1.301.@2 o -5.39 0 -1.872-4.56 0 0 0 -0.821

“%8 0“29 8.09 1.33 0.63 0 -1.6.630 -3.864.58 -2.S9 O 0 O 0.536

3B9 o 0 0 0 lS.91 -11.55-0.936-0.29-344.51.872 306.7 0 -306.78.9$

0 00 -0.936-IL9’7-2.28-1.301.872-5*39 o 0 0 -0.444

,;; 306:7 0 Q : Y&.; : ; o 306.7 0 -4907.20 0 0>W +)20.1 . 0 -%.7 o 4)07.2 -9347755.7 0

Qm -652.654.:5 0.1°H.1549.78LJ8-3.380.131.6~.0849-309.~0.45 0 755.67-6&6 71.84

~~ 6.35 19.50 -0.3140.498.30 10.55-OJJ 0.2648.98 -0.445 0 0 0 -224.04

+. co

i1 I

-.

.

50 NACATN ~?O.1014

.

TableVI. X Matrix.

%7 ‘B8 %9 ~Bt

37. ~957.92 18.91 0 -1226.878 18.91 -880.92 37.82 -107239 0 18.91

●

-344*5 0‘~Group 0 77.0 613.4 .-9047

TableVII. RingMatrixNo.1T137 %7 ‘B7 ‘B&3 % ‘B9 %9 *

-139.9 0 -I-6.63 53*45 -JJ.w 8.09 0.0 -5.39‘ o 11.97 -2.28 1.30 0 .

-16.63 0 -3.86 8.09 -1.30 0.63 o53.4513..97 6.09 -139,9 0 .4.6.63 -23.94 -

‘-3.l.97-2.28 -1.30 o’ -5.39 0 -f!+.568.09 1.30 0.63 -16.63 0 -3.86 -2.590 0 0 -11.97 -2.28 -1*30 -5●39

TableVIII.RinRMatrfiNo.2

xx T= %6 %677 -957.92 () 1.872 0U7 o -139.9 0 -16.63

‘7 L872 o -5●39 0W7 > 0 -16.63 0. -3.e3

.

.

.

.

.

.

. 4 n, ,

●

I

ILotione

~ 0.021@3

V6 0.47026

W6 0.35857~pwly O*W7W7-W717w+ 5*3yg o.llf3413U$-1.19199‘v&2.ao219W&2.12196)9 2.0422~’ 0.2MQ2~~ 0.07274w@.c05V97.3UH2

SIIlrJ

%5 ‘B5 %5

TableVIII. CheckTable

Udxilancesin W3. andin.lb. .

%5%6 %6, P

0.06628“ -0.0405 -o.c033 4.0010 -3*3575 o 0.c066 o’ e0.2479 1.u72 0.2569 0.1736 0 -3.0027 0 -0.3569-o.f@2 -5.6290,-1.olz! -0.6U3 0.ES03 o -2.5347 ()0.1040 2.WCEJ o.ti~ 0.2259 0 -5.%30 o -1.3W+J.

-o.17416 0.UX)38 0.c086 0.00271.664.0 7.7005 1.7245 1.I.6551.2329 15.7665 3.(X5L 1,71234.4602 -12.8375-2.0629 -0.9997

66.9281-66.92fAo’ O.O&@ o

66.92SJ.4.9283.

O.mll 1.766 O.om -o.(x)03O.oo1o

4.02202 -L5773 4.3s25 +.2U7 1.5513 1.811.1.o*.1449 o.1408I

f

Motions

Remits forcompleteQ@~ 6 0.~3505U6 0.021.463

V6 0.&7026W6 0.35857~, -0.0092J.y o.14407W/ -1.31717~ -1.58683~g O.w118-1.19199v8~.80219w~ 2.1219639 2.0422?* o.Z1822

~m 0.07274w~ .-0.005V9 7.3J.332

sum

maJ

~ble VTtI. CheckTable (contld.~

Unbalancesin lb. andim lb.

0.066284.2479-0.4402-oJo408.8224

0

-2.46570

2.2405-13.76752.6228

0.6M37

66.9281-66.92Jxl3.2633

0.04051.U.72

5.62902.9m8

o

-20.15540

26.3890-~.%a43.7U833.5JW17.1666

0.031.3

-o*m33AL2569-1.0722-0.466J.-o.Orla

o

7.09950

-o.llcq-1.4.26816.389o2.7585

dMoo5

O.oo1o0.17360.6u30.22590

-2.3959.0

6.1252-0.0344-9.64323.6428l:W

0.CO08

71.17

-0.17416.l.l%$o1.23290.4602

-1o4.37140

-5.24570

38.6Mw66.9281.-61.32713.9079

-0.10638

7.7W-15.7665-X2.8375

oU.67594

, 0-35.2882-23.5874

0.o169

#

o.oo841-1.7a53.0031.2.06290.221.8

0M.1038

o-1.9115

4MO07

-6.8453 *.54X -16.67u“

4M027

1.I.655-1.7U!34.99970

19.822ao

4.1908-0.5922

0.0C04-995073

0.6054 1.6110 0.05176 0.0439 2.6894 -4).6496 0.08914.01.63 z.

,.

. eI

.

Motions4 Resultsfor

Cc@ete &l.

>~ 0.003505U6 0.02U63V6 0.47026W6 0.358566

$,-0.00921

q 0J44073W/-1.31717q -1.5e&13$8 o.llww .-l.1919WV8_2.802185w~ 2.M196

79 2.0422

~~ 0.21.922

-f~ 0.,0’7274~B#Mo5

m) 7.s-332sum

TsbleVIII.OlwckTable(contld.)htdsnces in h3. snd in.lb.

%9 Xm x~,

613.32

4.480927.53505.%57.1.2307

-703953866.9281..22.30941.54791.3.6905

2.150

-5.649

-0.22M -.673

28.536212.7780

-5.49593.823 626.343

-1070.849356.950

-0.0006-399-

755.67

-4*W

11.299

-K7.01

-658.08-3.778

# &

MA, Yfi

7&

-4.561 0.02411.0903-.o.15s0.4590,

11.984 -o.11.p5,6158o.8q2-1.5%9

-w.% 1.%67-25.151.O2.3W1.1374

-626.343lf3.3390

54.970v 33.%8

-3.247J-

5.6700 0.0001 43.382 -26.199 29.598 1.5332

!’

——

Tabb IL F- l?elaxatimTable

Unbala&es in lb. andti.lb.

MDtion9 %4 T% %4%4%5 ‘B5 %35 %5ofTablem -0.02202.1.5773 -0.3s25 am?

.V5= 4.1 0.0935 1.1970 0.228 O.lgm 4.1872 0 0.539 085 = -0.08 4.0232 -0.&72 -.I.04-0.0% O 1.3304 0 0.3088V6 - 0.08 -0.0749 +.9576 -&m -0.~U7a 0.01* O.w L4128 O.om O.(X)O2o.0w8 o 4.0354 0 -0.0009ItlllalReEiduak1.4832 0.5825 o.u&2 0.08M -o.ay .1.2399 Q.lx)a -.@ml

MOts.cm

ofTeble VIII

q - -o.llmj * -O.(M

V’6 - O.CM

g :%

Find Reeidllel

%5 %5% % %7 %7 %7 %1.5513 1.8UI 0.1449 0.lA08 0.605b 1.6110 0.W8 0.0439

0.09364.1$7 0.228 -0.130.0232-0.6472 O.1.cu-0.05040.W8 ‘ o -o*4312 o 4;(77.490.9576 4.1824 O.wo.11550.53450.U.970.C809 o -L399 o -0.I.663.l.w?a-0.0327O.aw -0.0M8A.&l@ -0.0250OJXW+ 4.0336

0.5206 0.4686 0.1656 o.o# -2.OK)l 1.1446 4.1302 -0.019

Motiala %8 %8 %8 %8 %9 % %1 XE@I !!@!m ~~

of Table VIII 2.68% -0.6496 0.0891 4.CIJ.635.67WI o.~1 *.3Q -26.19929:5981.5332

Q7= O.o1 4.1155 0.5345-oJl$q 0JM09

&q= 4.aM 4.154 -1.224 19.629 MM% -L5D

~ .0.004 -3.1263 4.01.35 O.fxm% -o.om3 -1.W3 o.m5 3.022-26.646

Fiml Eesidl=J-8-0.7%4 ~.1= -0.03 0.0643 3.2Y3570.CXU)6 1.a7 -5.W3 1.4411.5332

.

I

,4.

—

.

.

m

NAOATN No. 1014

TableX. Complete(MrationsTableforLarg7eCylinder

Ringzil

AA

BB

cc

Ill)

u BB cc DD

1 2

3 4 2

3 5 6

‘7 8“J

.

TableXI. SectionL of(MerationsTableforLame Cylinder

.

Stringer

b

RingBB

FlixMBB

No. 1 2 3 & 5 6 7

1

2

3

4

5.

6

7

8

.

a c

bed

fed..

fed

fed

fed

fe

f

$9

d

55

●

—

. 9 c1 a

*’

56#

NACATN ~Oo 1014 ._ __,- —-

.

TableXII.Portionof@ erationsTableforLargeCylinder .

Ringcc I tRj.ng Dl)StringerNo.1 234567$9 123456789

●

lac I TV2 bed

1“Uxw

3 fedI

Yxw

4 fedI

Yxw

5 fedRingcc

6 fed

7 fed

8 fhj

Yxw

Yxw

Yxl’r

yrs .tu

.

lTVJ HK

2 utx~ JML

3 y! ~ ;1 NML

4 y!p Wt NML

5 yfx Wt NMLRingDD

6. y!x w? NML

7 yt x w? NML

$ yf ~1 *I Nmn

9 tt~ Pq.

.

.

.

.

.

.—

.

NACA TliNo. 1014 57TableXIII. Elementsof Sectionsof OuerationsTableforLargeCylhder

. X.T’R

Elementa

N

1.3

7.;11.280.625

-:.99

-3%?4

7.$-1.280.625

. x T R N

Element c1 . .37.82o“ -1140 A

.y -689.0- 0

-5.49v o

Element b . Elementg

?5.63 do 22.8

“o -;.$o’ ●

.Element c Elementk (cutout)

o 0“U.40 -2.24

18.91-IL55 -0.936-0.936-IJ..97-2.28.4(

Elementj (cutout)

-0.29-1.2$%

Elementd

37.82 -1.872-23.1 -23.94-1.872 .-4.567-0.58 -2.592

46:4.:.4031.40-2.247.91-1.28

Elementh (cutout)Elemente

. ~ -689.0u ov ow o

-631.6~1.;5 0.936U..55 O.y?l

. 0.9360.571 -5●540.29 -16.810.01.4

0.29-16.810.034-3.88

0-1;4.0o

-5649-1:.99

FornocutoutElementf

$ 37.82u ov ow o

o 046:4 11.40-11.40-2.247:91 1.28

Elementr (cutout>

L872-5.39

“Elementm (cutout)

. 3 -823.64$.:50;9:;: ---$5

13.65 -2:698. w ~ 29 -8.32 1.972

-Q.29-8.321.972-1.932

Fornocutout

S=zt =-VIu.TForCtiwt

Elementr!S.t. u.oS:=t!=olz=puq=o , -0.9360.29

~.8710.177249.3y*:;-U-550.936-0:571-0.290.177

uvm.

0.0462-0.OU4.0L!4D.0044

—

%‘4‘A

:5w;

%V6W6

,. ‘9

\

TsbleXIV. RLnalDisplacements for harp-eOyIinder

Ax&l misplacements %in ●M1 in.

Ri.ngA RingB Mngc MD

31 ,1o.47k2E) 0.;236 : 0.19605 0,05024

0.02706&.078~ ;“~3; -0.03364o.021k5 . o.08m0.22974, : 0.47192 OJ.86281.40256 3.3QJ3

~ gllammmts in limes of ;RUl#& ill .WL in. or .001 rd.

7.6981 3.849

0.0;0.04

0.036150.114580.03033

0.03642-0.45799-0.47259

-0.36193-0.634950.76479

A034051.73$!35-0.4733B

-1.48075

1.2830

-oro50200.362600.66364

0.3741Jo.55414-L01502

-o.22r74-2.m550.99654

-0.321.58-0.58833-2. 15@6

-2.I.6234.775413.30103

13.35862

0.01-0.05-o.12

0.373852.3921.30.59187

0.51767-5.379m-44485

-3.23801-:.~544

●

5*329U53.2072517.21916

El%

‘1 I1’;’

j; I!’ I

. b, , r b

Tsble~. ~S idue@ forLarJzoCylinder

AxialForcesin lb.

B6B7 B8 B9 C6 07 C8 09

1.I.86-U.687 -8.6% 13.493 6.3u -5*775 4.514 -11.015

D5 %

-5.287 -10*Q2

q Rh

1.IQ14 O.11’p

1.W954.U82

o.1882 o.02fJ

& T’70.0470 -0.0208

4.0297 4.+5

-Q.02JJ5-1.898

D’? D8 Xm

-3.088 13.169 21.47

RQ B - FOI’I%S in lb.

N4 T5 %

0.0403 1.8386 -o.lXp6

RingC - Forcesinlb.

-0.05090.7087 0.0572

~ D - FUrC13S in lb.

4.0U$3 -1.07530.1361

RingB - Fm’cesin lb.

% 9 T8

-0.0488 0.0337 1.094

RingC - Forcosinlb.

0.0960 -0.0294-1.957

~ D- Forcesin lb.

-0.0275-0.0252-1.7565

%3 %D

9.% -8.76

Wmsntsinh. lb.

, N5 %

0*0381. 0.5852

Momentsin in.-lb.

-0.0440 0.1.126

Monmts in in.-lb.

o.0219 4.0938

Momentsinb’1.-lb.

% h

“o.1464 0.0376

llomentsinin.-l.b.

-0.0864 -0.0501

Momentsinin.-lb.

-0.I.682 4.0443.

~,

0.01

%

4.U27

-0.0384

-0.0196

%

-0.3491

0.L!#54

.

I

.

.

.

.

NACATN No.1014 Fig. I

(a) ACTUAL CYLINDER

! r “ 2’7

.

1

1r --i YA 2011

rB,~B ,

1 .1I I II 1- 1 7 @ 6.43 = 45” ~

i’{~ ~34~—

+zl3/. EZZZT g

1-STRINGERSECTION

RING SECTIONA-AB-B

(b) SIMPLIFIED CYLINDERA=0.093SSQ.IN.A=O.1877SQ.IN. - X,5 ‘

Bc ●

D

.

..4

P-A=O.37SSSQ.IN.~ 12.86”d+.64lt-

! ,—

1

-1-—

I )M,

RING PLUS EFFECTIVESHEETSECTIONC-C

FIG.1. MONOCOQUECYLINDERSHEET: 24 ST ALCLAD

REINFORCEMENTS‘ 24 ST ALUM.ALLOY

Figs.2,3 I ,NAGATNNo. 1014

-%

/’

-+

L/ r

\

.I t--a--l

1/

f———/ I

I %6

/

i

——— — i

FIG.2 UNITOF A

I

(\N /R

BEAMCONVENTION

R

Nf\

L

AXIAL DISPLACEMENTCORNEROF A PANEL

\NT(/R

1

FRAME

)

CONVENTION

R\N{

[/

T

.—

.

.4

.

.

.

.

FIG.3 SIGN CONVENTIONS

NACATN No. 1014 Figs. 4,5,6

.

.

D

c

A

4

L

B’B,C

.

FIG.4 UNIT TANGENTIAL DISPLACEMENT

A

E

)’-

t ..

q ‘vt. !——— ——

B,C

FIG. 5 UNIT RADIAL DISPLACEMENT.

A D————

I ‘1.

1

I I——— ——.

cB)C.

FIG.6 UNIT ROTATION

—. ---

.

Figs. 7,8 . ‘NACATN No. 1014FORCESAND MOMENTSACTINGON CONSTRAINTS.FORSIGNS S= FIG.10a,

~J a=@—~K aI,lX 1

FIG.7. EFFECTOF UNITAxIALDISPLACEMENToF F.r=.!&

—

—

.

.

&

a

.

.

FORCESAND MOMENTSACTINGON CONSTRAINTS.FORSIGNSSEE FIG.10a.A B ,

m I IE

FIG.9.‘EFFECTOFUNIT RADIALDISPLACEMENTOF F.p=y A=+

FORCESAND MOMENTSACTINGON CON~RAJNTS.FOR SIGNS SEE FIG.10a.A c

I

:

ii

Figs.Ioa,1:3,14.

NACMTN No. 1014

G

FIG,10.a, SIGN CONVENTION FOR--FIGS. 7-10.

SIMPLIFIEDCYLINDER

FIG.13.DEFLECTIONSOF CUT RINGPURE BENDING.M= 6155.4 IN. LB.

FOR

●

●

—

- - ,-. .....-

.SIMPLIFIEDCYLINDER

,—.. — A—y—

FIG.14. DEFLECTIONSOF COMPLETEPURE BENDING.

.

RINGFOR a

M = 6155.4 IN.LB.

.,

‘, b,

10 I 1 I I I IEDGKOf’ WITW ~ SMPUFIED CYLIND~

- - - --—. —

a

7 \

603MPLm- RING

sCUT_

$Z4 RING \u~

-3 ‘ k

E1-1112

351 \

J20 . ~- ~z -r 4- 20 m .2 .4 .6 s la

.

I I

.ODi’I I

SIMFUFIED CYLl?4XR2

/ “

0 K

/ ~

A

-2

I

_ I0 L —

— _ _B . --/ ?

-1

2

0 M —~ c G

-2

2

0 H - — -, ‘i_ ‘ ‘ ~ — . H

-2I

*I-

0 0- ~

EL \

1 . J

-1 ;

.

,+WD 0X4FUTEEND RING k?G

FIG. 12 RADIAL DEFLECTIONS FOR S+RINGERS ~

DUE TO PURE BENDING.In~

M ● 6155.4 N. ~.--N

POSITIVE SIGN lmlCATfS INWARD DEFLECTION

,.-FIG. 11. AXIAL DEFLECTION OF RINGS

PWIE BENOIMG M = 6155.4 M.LB.I ,

—

1’ FIG. 15.

‘11.;,

A)(l& DEFLECTION OF RIMSCOMPRESSION P = -1286 LB.

Ir, b

‘1-

TmoENO

mwLETE CUT ~RING RING k

FIG. 16. RADIAL DEFLECTtOPdS FOR STRINGERS ;

DUE TO PURE COMPRESSION.P= ‘1286 LB.

,5

PIWTIVE SIGN INOICATES INWRD DEFLECTION 5

F

\

., b*

SWPLIFIED CYUWR

FIG. 17. DEFLECTIONS OF CUT TUNG FOR

PURE COMPRESSION.

I P --1286 LB.

,1

I

I. r h.. ,

SWPLIFIED CYLINDER

.

FIG. 18. DEFLECTIONS OF GOMPLETE RING FORPURE COMPRESSION.

P=-12S6 LB.

4z

.

-n-.Qm

“:m

I

10EDGEOF UJ7W7 ‘---— EXFHWENTAL—. — .— .-

9 /’ ;‘—— RESULTS

(I

!

SIMPLFED~, \

—-——

8CYLINCER

\\\ \.

i

7cY’’N%R

\“

\“ , ‘\6 ) * \

\\ \

~s

:04g

~-3

k:2

< \y i\al

d$~

2 \o \’ ~ 2 PN \ \ ST RESS x IO-J~1 9\\

\E \. \

$\

11.3

g4 $! .1- !

~ \

;5\ \

6t

7

!

81

\

9i

1 FIG19Q

NORMAL STRESS DUE TO PURE B;NDINGCOMPLETE FIELD ,.

M=35@0 INIE.

I01 I IuxF. mwTwJT ------ EXPFRI-AI I T-.

g

9i . . __ —..-. ..... . .. _

‘—— RESULTS

\ —;-.—s&~m~&D

8

cnlmER4

\\ I 1 I 1 I;2 - I I I I 1 I 1 ,

$?:

FIG:20. NOIWAL STRESS OUE TO P& BENDING -CUT FIELD.M = 35$)00 IN. LB.

i: !,..> ,

iv ,. a v, *-

,1’i11’.’. .[1 I

. b +

10

9

8

7

6

‘6

7

8

9

10

,EOGf&F+W7 :yklm:

/

K L 1

I /

-7 -6 -5 -4 -3 -6-5 -4 -3 -2 -t aSTRESSX IU2P.S,L STRESSX10-2P.SJ.

CUTFIELD COMPLETEFIELD

“NORM!4LSTRESSESDUETO PURECOMPRESSION

UJ .—— ——. —. P

?Pm

+

I

(-$

IIIII1“ItIII

.— ——— . ——(J

Iv“N“NNcd

P = -1286 LB.FIG.21 FIG.22

.10 1

EDGEOF CUTWT ~

9 ‘---FIELD AB ~

8-

-i

7

6 ,i

,5

4 .,

3 “ /

2

I \

0 -8-7~* -5 -4 -3 + -+’\

I Smss x10-~P.s.l.II -

2

3 .

4

5U-A

S?@LFED CYLNEER

6 . —.—~E FIELO—— —CIJT FELO r

1 FIG. 24. SWARM STRESS W TO PWUZ ~lKM = 35,000 IN. LB.

10 I I 1 1

l-. ~EDGE OF aNCur

-----9 I I I I I - I 1

6 \ )

w_FELO

7 / ,’/ / CmWmT

\ / —, FIELD6

5 -I

\

4 \

3

2 // ‘

I

o [to -60 -4 0 0 20 40 6060

S7RESS P.s.l.I

2 \

3

4 L

5

6

7[ SMJLIFIEO CYuNbERIllil

10I I I I I I I I I J

FIG. 25. SHEARING STRESS U TO P(K CCNPRESSDNP = +266 LB.

NACATN No.1014

.

●

.

●

4-RING C-C RING D-D

~3 %.

\\

:2 ‘!!.

k\

aZ 1l-.m% \ w

~ ~~ ~

~ COMPLETERING

SOGE OF’ CUTOUT

-3-- I I0 10 20 30 40 50 60 70 60 90 100 ]10’ 120 J~ 140 150 [60 170 [60

z; DEGREES

FIG.26. BENDING STRESSES IN RINGSPURE BENDING M s 35,000 IN.LB.

SIMPLIFIEDCYLINDER –>-- -- LARGECYLINDER

*>\r,\ 21 ‘

3’ \\

AEW.-0.18774, SQ. IN.

\

\,.— -—- -51

,/

>’q3/3/

/ ,0’ 4

///.- —. —-

5

.

“Em=l.

L6.43d-6.}3.&6.4311fl

)M

FIG.27. SMALL CYLINDERSHEET: 24 ST ALCLAD

REINFORCEMENTS: 24 ST ALUM.ALLOY

Figs,28,29

-.<.

NAGATNNo.1014la

9

--- 0

7

6

6

7

8

Q

Ic

.

\ .

/ \

\.

J 4! ( 2 3\ STRESS X 10-3 P.~:l.

\

CUT FIELD~

$

-—.

COMPLETE FIELD~

EDGE OF CUTOUT

FIG.28. NORMALSTRESSESFOR SMALLCYLINDER.M= 42000 IN. LB,

*

.

3“~22 Ix~~ , L

gg+ao~g -..

~rn \-#J-l- ,~ \; -2 I I

.-.● EDGEOf CUTOUT

-30 10 20 30 40 so 60 70 60 90 100Ilo120130140 150160170 rob””””-

—..

. “J

-.

1

i’..

.,

L

..-. .—

*

b

3.

_=. -

,* d, DEGREES

FIG.29. BENDING STRESS IN RING I=OR SMALL cYLINDER.M = 42000 IN. LB.

.

NAM

●

TN No.1014

..

Figs.

10

!3‘

8 -

7

6

#x04~

& 3

E:2

alJ~.

-4 -3 -2 -1 0 1234R~1 STRESS X 10-2 P.S.I.

%

$2.

k3ld24 \

g5

6

7 /

8 /(

9 .- — .— - y .—. -— - -— -.EDGE OF CUTOUT - T ~

FIG.%0.SHEAR STRESSFORSMALL CYLINDER.M=42000 IN. LBm

30,31

FIG. 31. NOTATIONFOR LARGE CYLINDER.

10+DGc w cm-our

8 -

7ExACT, -APP Roxl MN-E !

6

501

“:4~

-3a ~W

‘2

!!al-1$0z - 3$jEl

?2

1! 3

UU4zg .

:5

\,6

7

? ,a

e

‘a FIG. 32 SOLUTIONS “OF THE SIMPLIFIED C~lNDERCOMPLETEFIELD

M= 3$000 IN. LB.

10 I -nGcffcuTwT

--.

‘- “ - -.- —. Q

p

cd

a~.

1 uw

7Ap Pmixl htnTG

6T

3564 T~

-3KL2

3alJSoz 3 2 -1 0 I 2 30!!!1

STRESSx 10*

$2 .’

~3

wV4z~250 z

6DoD

7\ ,

+z

8z.0

QG

\;F

‘0 FIG. 33 !X)LUTKINS ~O& WLDSWLIF.lED CYLINDERJ

4

*. 7’ Fw .!

, L -4 w%

1! I I-. .